A) Find [Zn^2+] and [Fe(CN)6 ^4-] in saturated solution of Zn2Fe(CN)6.

Question

B) Repeat the above question in 0.1 mM ZnSO4 saturated with Zn2Fe(CN)6.

Answer 1

A. The dissociation of Zn2Fe(CN)6 is:
Zn₂Fe(CN)_6s --> 2Zn²⁺_aq + Fe(CN)₆^4-
For Zn2Fe(CN)6,
Solubility = s = 3.74x10^-6 M and
K_sp=2.1x10^-16
[Fe(CN)₆^4-] = solubility
[Zn²⁺] = (2)(solubility) <--- see equation above

B. K_sp = [Zn²⁺]²[Fe(CN)₆^4-]
2.1x10^-16 = (0.1)²x
Because of the low solubility we assume that the added concentration of Zn²⁺ is roughly equal to the total Zn²⁺ concentration. "x" represents the ferrocyanide ion concentration which changes because of the equilibrium shift caused by the addition of Zn²⁺ cations. The approximate concentration of Zn²⁺ is 0.1. To get the concentration of Fe(CN)₆^4- solve for x